What Happens When a Gyroscope is Spun on a Horizontal Surface?

[bobie]

Please , watch this girl ( at about 50 sec) ,
can someone tell me what happens if the wheel gets a spin when it is already on a horizontal position?
Thanks.

 

[Bandersnatch]

If the total angular momentum(in the horizontal plane) after the initial spin is some value L=L₁+L₂, then after flipping the wheel vertically the system of girl+wheel has got to have the same value of angular momentum L=L₃+L₄

Let L₁ and L₃ be the ang. momenta of the wheel, and L₂ & L₄ of the girl. Plug in some random numbers and see what will happen. Remember that L is a vector, so a negative value means spin in the opposite direction.

Try to visualise what will happen BEFORE watching this video(at 5:00 minute mark) to check your understanding:
https://www.youtube.com/watch?feature=player_detailpage&v=sy5NY-Dqdys#t=300

 

[bobie]

then after flipping the wheel vertically

I mean no flipping or tilting: you give a horizontal spin, (or hand over the gyroscope to the girl with the wheel spinning horizontally), and leave it like that.

 

[bobie]

after flipping the wheel vertically the system of girl+wheel has got to have the same value of angular momentum

The title of the video is appropriate only for the first part of the video.
In the case of the-girl-with-spinning-gyroscope conservation of L is not relevant.
What applies here is the 3rd law of motion: the turntable spins the same way as it would if the girl threw a ball tangentially or tried to run.

 

[Bandersnatch]

The title of the video is appropriate only for the first part of the video.
In the case of the-girl-with-spinning-gyroscope conservation of L is not relevant.
What applies here is the 3rd law of motion: the turntable spins the same way as it would if the girl threw a ball tangentially or tried to run.

Of course the conservation of angular momentum is relevant.

If you set up a system with a given total L, it will remain equal at all times(barring frictional losses).

If you hand the stationary girl a spinning wheel, oriented in any direction, you set up the system to have the angular momentum equal to 0(girl's) plus L_w(wheel's). The total L is then equal to L_w.
The girl can not start spinning then, as that would require L_w to change, and it's got no reason to.
That's pretty much the situation depicted at 5 minute mark in the Sixty Symbols video. The man has stopped spinning due to friction, and now he's holding a wheel horizontally, just as if it was handed to him that way. The wheel is not making him spin.
It's only when he tries to flip it, the change in L ensures he starts spinning.

If you try spinning the wheel while the girl is holding it, then of course you'll apply torque from an external source, that will make the whole system rotate. Both since the torque is applied in one direction only, the whole system will rotate in the same direction.

Once there are no external torques present, conservation of momentum applies.

 

[bobie]

That's pretty much the situation depicted at 5 minute mark in the Sixty Symbols video..

In your video, (exactly at 4:28 mark) the wheel is handed over to the man spinning vertically, the same as to the girl in my video, I can't see why you posted it.
I think you are hinting that if the wheel were handed over horizontally, the turntable would start to spin only when the man/girl tried to tilt it to a vertical position, is that correct?
Are you saying also that the turntable would rotate if the wheel should be given a spin while (at rest) in the person's hands?

 

[Bandersnatch]

Yes to the first question. Once again, it's depicted in the video at 5' mark when the man is seen sitting still with the wheel turning horizontally(it doesn't matter what happened before, as at this moment he's stationary). It is exactly the same situation as if you handed the wheel to him.
The fact that it is spinning does not create any torques by itself. You have to tilt it.

And yes, the turntable would rotate, but not due to the conservation of angular momentum, only due to the application of torque. Somebody outside the system trying to spin the wheel would act to both spin the wheel and the turntable at the same time. Since the torque is applied in the same direction for both elements, both would spin in the same direction.

But, if the person tried to spin the wheel by him or herself, there would be no external torques, and the conservation of ang.momentum would apply. Meaning, the person and the wheel would spin in opposite directions.

 

[bobie]

Somebody outside the system trying to spin the wheel would act to both spin the wheel and the turntable at the same time. Since the torque is applied in the same direction for both elements, both would spin in the same direction.

Watch this video, http://www.fizik.si/index.php/en/fizikalni-eksperimenti/vsi-eksperimenti?start=40&videoid=-Cc-jGnIwCM
at the beginning someone outside the system spins the wheel and the turntable (does not move, but) spins only when the wheel is twisted.

 

[Bandersnatch]

Watch this video, http://www.fizik.si/index.php/en/fizikalni-eksperimenti/vsi-eksperimenti?start=40&videoid=-Cc-jGnIwCM
at the beginning someone outside the system spins the wheel and the turntable (does not move, but) spins only when the wheel is twisted.

Yes, but it's spun when it's vertical, so the torque appplied is neutral with respect to the turntable(chair) axis. In post #3 you said you wanted to know what happens when the wheel is spun when held horizontally, which is what I addressed.

 

[bobie]

If you hand the stationary girl a spinning wheel, oriented in any direction,...The girl can not start ... (1)
If you try spinning the wheel while the girl is holding it (horizontally), then of course you'll apply torque from an external source, that will make the whole system rotate. Both since the torque is applied in one direction only, the whole system will rotate in the same direction. (2)

Is that correct?, what do you mean by "in one direction only"?
If you could recap, please:

1) - A spinning wheel is handed to the the girl horizontally (no turntable spin)
2) - the horizontal wheel is spun tangentially by someone outside the system (turntable, of course, spins same direction of the push...)
... but what happens if the spin is applied in a radial direction toward the centre of the turntable?
3) - then the girl tilts the horizontal wheel on a tangential plane by 45° or 90° clockwise (...?..), anticlockwise (...?...)
4) - does the degree of the tilt affect the velocity of the spin of the turntable?
b) I suppose the velocity changes if the girl moves her hands radially
5) - if the girl gives only one tilt (and leaves the wheel at 45° or vertically), the turntable spins for a while and then stops due to friction, right?
Can we conclude that if you do not tilt a gyroscope tangentially the turntable does not move?

 

[bobie]

A couple of more questions:
when we tilt a gyroscope do we slow down its spinning? do we substract Ke?
what happens if the girl tilts the axis horizontally, (like a bicycle hand)?

 

[rcgldr]

when we tilt a gyroscope do we slow down its spinning? do we substract Ke?

Angular momentum is conserved, but internal work can cause an overall change in energy. For a simple example, when a spinning ice skater pulls in their arms, work is done, and the energy of the spinning skater increases, while angular momentum is conserved.

 

[bobie]

when a spinning ice skater pulls in their arms, work is done, and the energy of the spinning skater increases, while angular momentum is conserved.

When a skater does that it alters the very parameters of L, if we tilt a gyroscope we are not modifying direcly L, I am asking if the indirect action of moving the gyroscope around can indirecly influence L

 

[Nugatory]

When a skater does that it alters the very parameters of L, if we tilt a gyroscope we are not modifying direcly L, I am asking if the indirect action of moving the gyroscope around can indirecly influence L

If you change the orientation of the gyroscope, then you are influencing L, and there's nothing "indirect" about it. The angular momentum L is a vector, not a scalar, so it has both a direction and a magnitude; changing the orientation of the gyroscope changes L by changing its direction.

Because L is a vector and torque is defined as $\frac{dL}{dt}$, it follows that torque is also a vector and has a direction. When the torque vector is in the same direction as the angular momentum, it changes the magnitude but not the direction of the angular momentum; the rotational speed about the axis increases or decreases. When the torque vector is perpendicular to the angular momentum vector, it changes the direction but not the magnitude of the angular momentum; the axis of rotation changes. Any other torque vector can be rewritten as the sum of the parallel and perpendicular components.

 

[bobie]

When the torque vector is perpendicular to the angular momentum vector, it changes the direction but not the magnitude of the angular momentum; the axis of rotation changes. Any other torque vector can be rewritten as the sum of the parallel and perpendicular components.

What happens if the girl tilts the gyroscope horizontally, right hand in, left out? will the platform rotate? if it doesn't does, the girl do less work?

 

[Nugatory]

What happens if the girl tilts the gyroscope horizontally, right hand in, left out? will the platform rotate?

When the hands apply forces to the gyroscope housing, there will be equal and forces acting on the hands. From this you can write down the torque that acts on the girl. If that torque is oriented in the right direction it will cause the platform to spin. What is that direction?

if it doesn't does, the girl do less work?

If the platform doesn't start spinning and the gyroscope maintains the same speed, then the total energy of the system will be less than if the platform does start spinning. So if energy is going to be conserved, the girl is going to have to do as much work as needed to get the platform spinning.

You can check this result against the formula that you've been ignoring for two threads and and what feels like a few hundred posts now, for the work done by an applied torque - the change in kinetic energy should come out equal to the torque times the angle over which it is applied.

 

[rcgldr]

If the platform doesn't start spinning and the gyroscope maintains the same speed, then the total energy of the system will be less than if the platform does start spinning. So if energy is going to be conserved, the girl is going to have to do as much work as needed to get the platform spinning.

If resulting torque on the platform is perpendicular to the platform axis, I would assume that the applied torque ends up affecting the Earth by a very tiny amount, and that the total energy and momentum of the system including the Earth would be conserved.

 

[Nugatory]

If resulting torque on the platform is perpendicular to the platform axis, I would assume that the applied torque ends up affecting the Earth by a very tiny amount, and that the total energy and momentum of the system including the Earth would be conserved.

Yes.

 

[bobie]

You can check this result against the formula that you've been ignoring...
equal to the torque times the angle over which it is applied.

I am confused, Nugatory: in the video the girl is pushing the wheel to her left, and the platform moves to her right, wiki says that the wheel has inertia and resists any change of direction, so, I thought that the platform is moving as a reaction to the force applied to the left (3rd law).
I started the new thread to learn how to calculate this force of inertia and consequently the force required to win it. You said:

If you actually change the orientation of the spin, you're doing work because you're applying torque across an angle, and that does work just as applying a force across a distance does. Indeed, it's fairly easy to derive the formula for the work done by a torque from the the old classic $W=Fd$ ] (and I strongly suggest that you try doing that as an exercise).

but everybdy said that the work done is (next to) zero

I'd really be grateful if you could be so kind as to help me with that exercise, as I am totally bewildered:
- the wheel has (m = 2 kg, r = 0.318 m, ω = 5 rps, v = 10 m/sec) L_w = 6.4 J*s
- the girl-platform system has (m= 50 kg, r = 0.4 m, ω = 1/2π rps, v = 0.40 m/s) L_g-p = 8 J*s
- the angle of rotation is 90° = 1.6 rad, d = 0.64 m
Is L_g-p =F*d?
is F then equal to 8/.64?
Is work done in order to rotate the wheel 2*8/.64 or zero?

 

[rcgldr]

I added this comment to the video, noting that there is an external torque involved.

"no external torque" - When the girl applies an internal horizontal torque to the wheel, the equal and opposing torque that would tend to rotate her body horizontally is opposed by an external torque due to a lateral force from the floor (surface of the earth) that is exerted onto the bottom of the rotating platform and the girls feet. The closed system needs to include the Earth in order to be a closed system. If the girl was hovering while in free fall inside an aircraft simulating zero g, then the girl and the wheel would be a closed system (ignoring tiny effects like aerodynamic drag). In this case the girl's body would be flipping and twisting while twisting the spinning wheel. In an ideal case with no losses, then the original direction and magnitude of the angular momentum of the girl and wheel would be preserved. Total energy would also be conserved, as any internal work performed by the girl would result in a corresponding change in the kinetic energy of the system (girl and wheel).

The point here is that the external torque due to the lateral force from the floor at some distance from the center of mass of the girl and wheel equals the torque being applied to the wheel. This external torque has a horiztonal axis (parallel to the floor), and the precession of the spinning wheel results in a perpendicular reaction to that torque, which in this case is a rotation reaction along a vertical axis. Using right hand rule, in the initial state, the spinning wheel has angular momentum vector pointing to the girls left. If the girl applies a clockwise torque, then the external torque is a clockwise torque, and the torque vector is horizontal, pointing away from the girl. The cross product of these two vectors results in a rotation vector pointed downwards, so the girl and wheel will rotate to the girls right. If the girl applies a counter clockwise torque, then the girl and wheel will rotate to the girls left. The angular acceleration about the vertical axis only exists while the girl applies a torque to the wheel.

What's notable is that in the initial state, there is zero angular momentum about the vertical axis, but after the external torque is applied, it becomes non zero. If the girl was in a simulated zero g environment, such as an aircraft or space station, then angular momentum is conserved, the direction and magnitude of the angular momentum vector would not change no matter how the girl twisted the wheel, and as mentioned, the girl would be flipping and rotating as a result of applying an internal torque to the spinning wheel. It would be similar to the twisting techniques used by divers, gymnasts, trampoline competitors, ... .

 

[bobie]

Thanks rcgldr,
that is very interesting but a bit too complicated/abstract for me, could you please translate this into practical numbers and equations with the (plausible?) data in my previous post, considering a concrete exapmle?

Thanks a lot for your understanding

 

[rcgldr]

The torque applied by the girl to the wheel is the same as the external torque applied by the floor to the girl which is why her body does not rotate about a horizontal axis. The platform, girl, and wheel are a complex system. The math would involve using the external torque and the angular momentum of the system to determine the rate of precession.

The zero g environment would change things. The initial angular momentum vector of girl and wheel points to the girls left. No matter what she does, that won't change. If she applies a torque to the the wheel, she "tilts" and and twists along with the wheel but the systems' angular momentum vector will remain the same.

When divers initiate twists while flipping (with no initial twist generated before leaving the diving board so that the initial angular momentum vector is horizontal and to the side of the diver), they will get tilted until they stop the twisting motion.

 

[bobie]

How do you relate the two values of L : 8 and 6.4 J*s?

 

[rcgldr]

I don't understand. In the initial state, only the wheel has angular momentum, the girl and platform angular momentum is zero.

You mentioned work done. Work done would be torque times angular displacement, and would result in a change in the kinetic energy of the girl + wheel + platform system if there were no losses. The issue is that the work being done transitions between positive and negative as the orientation of the wheel changes.

 

[bobie]

I don't understand. In the initial state, only the wheel has angular momentum, .

The platform acquires L = 8 J*s, isn't this related to L_w? if L_w had been , say, 20 instead of 6.4 wouln't it be different?

.. when the girl peforms work by exterting a torque over some angle,..

What is concretely the value of work in this particular example, and how do I find it? don't I need to relate the two values? that is what I still do not know, how to calculate it in a concrete example

 

[rcgldr]

Another issue is that in addition to applying a torque that forces the wheel to rotate about a horizontal axis extended outwards from the girl, the girl also applies a torque to prevent the wheel from precessing about a left to right horizontal axis in front of the girl, by keeping her arms straight and not allowing one arm to move inwards with respect to the other. This torque would also correspond to an external torque related to lateral force applied by the floor onto the platform onto the girls feet.

 

[bobie]

the girl also applies a torque to prevent the wheel from precessing about a left to right horizontal axis in front of the girl,
. This torque would also correspond to an external torque related to lateral force applied by the floor onto the platform onto the girls feet.

If I got it right, when she pushes her right hand down she meets no resistance but at the same time the right axis starts pushing horizontally (clockwise looking from above) and she does some work pushing anti-clockwise leaning on the platform that starts moving horizontally clockwise as a reaction (3rd law).

Is that right?
If so, a couple of questions:
- can the girl modify the speed of the platform in any way (for example) tilting the axis faster? or is the outcome determined only by L_w?
- can the situation in space can be obtained hanging the girl from the ceiling?
- is L_g-p 1/2 the work done by the girl or it follows the rules of elastic collision m*v=m1*v1?

Thanks a lot for your invaluable help

 

[rcgldr]

In the wheels initial position, the girl inputs a roll torque. She also inputs a yaw torque to prevent the wheel from precessing about a vertical axis with respect to the girl by keeping both arms extended. The girl will feel a resitance to any attempt to rotate the wheel along any axis other than the wheels primary axis, and this resistance will be greater if the wheel is spinning faster.

To simulate the zero g situation, the girl could be harnessed into a gymbol type apparatus to allow rotation along any axis, similar to ones somtimes seen at carnivals, but it would have to be made of very light materials, perhaps one built from carbon / kevlar fiber (the carnival ones are metal and somewhat heavy, too much inertia).

 

[bobie]

a resitance to any attempt to rotate the wheel along any axis other than the wheels primary axis, ...).

Thanks, rcgldr,
I know that after 40 years these problems were raised , some processes are still mysterious, do you know of a reason why the wheel stops dead although it's on frictionless gimbal (here at 1:45) and why the big one is so light at
3:20?
BTW he says at 1:50 that there is no angular momentum, isn't it inertia that is missing?